How to solve $6^{2x}-10\cdot 6^x=-21$ using logarithms?

Question

What do I do with $\large 6^{2x}-10\cdot 6^x=-21$?

Since $6$ and $-60$ are not of the same base (nor can they be written as exponents of the same base cleanly) I am having trouble solving for $x$.

Answer 1

Let $6^x=y$ $$6^{2x}-10\cdot 6^x=-21$$ $$y^2-10y=-21$$ $$y^2-10y+21=0$$ Factor. $$(y-3)(y-7)=0$$ $$y=3, \ 7$$ Replace $y$ with $6^x$ $$6^x = 3, \ 7$$ I will solve for both equations separately. Let's start with $6^x=3$ $$6^x=3$$ $$\ln(6^x)=\ln(3)$$ $$x\ln(6)=\ln(3)$$ $$x=\dfrac{\ln(3)}{\ln(6)}$$ Now for $6^x=7$ $$6^x=7$$ $$\ln(6^x)=\ln(7)$$ $$x\ln(6)=\ln(7)$$ $$x=\dfrac{\ln(7)}{\ln(6)}$$ The solutions are: $$\displaystyle \boxed{x=\dfrac{\ln(3)}{\ln(6)}, \ \dfrac{\ln(7)}{\ln(6)}}$$